Codeforces 545D Queue【贪心+模拟】

最新推荐文章于 2020-07-21 22:49:16 发布

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最新推荐文章于 2020-07-21 22:49:16 发布

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分类专栏：贪心文章标签： Codeforces 545D

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本文介绍了一个算法问题，即如何通过调整排队顺序来最大化顾客满意度。具体地，通过将顾客的忍耐时间排序并模拟排队过程，实现了求解最大数量满意顾客的算法。

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Little girl Susie went shopping with her mom and she wondered how to improve service quality.

There are n people in the queue. For each person we know time t_i needed to serve him. A person will be disappointed if the time he waits is more than the time needed to serve him. The time a person waits is the total time when all the people who stand in the queue in front of him are served. Susie thought that if we swap some people in the queue, then we can decrease the number of people who are disappointed.

Help Susie find out what is the maximum number of not disappointed people can be achieved by swapping people in the queue.

Input

The first line contains integer n (1 ≤ n ≤ 10⁵).

The next line contains n integers t_i (1 ≤ t_i ≤ 10⁹), separated by spaces.

Output

Print a single number — the maximum number of not disappointed people in the queue.

Examples

Input

5
15 2 1 5 3

Output

Note

Value 4 is achieved at such an arrangement, for example: 1, 2, 3, 5, 15. Thus, you can make everything feel not disappointed except for the person with time 5.

题目大意：

一共有N个人要排队，对应每个人如果等待的时间超过了自己的忍耐时间，那么对应这个人就会走，问怎样分配这些人进行排队能够使得最多的人得到服务。

思路（这个D好JB水啊）：

直接将每个人的忍耐时间从小到大排序。

接下来我们模拟整个排队过程即可。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define ll __int64
ll a[100060];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=0;i<n;i++)
        {
            scanf("%I64d",&a[i]);
        }
        int tmp=n;
        ll sum=0;
        sort(a,a+n);
        for(int i=0;i<n;i++)
        {
            if(sum<=a[i])
            {
                sum+=a[i];
            }
            else
            {
                tmp--;
            }
        }
        printf("%d\n",tmp);
    }
}